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I have two equations

$$\dfrac{\dfrac{P}{1-(1+\frac rn)^{-n}}}{\frac rn}=x$$

and

$$\dfrac{P\cdot\frac rn\cdot(1+\frac rn )^n}{(1+\frac rn)^{n}-1}= x.$$

If I put in an amount for $P$, $r$, and $n$ I get the same answer in each formula but I cannot figure out how to simplify the second one to see if they are truly the same formula in different formats.

Xander Henderson
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Molly Jarrell
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  • I have edited your question , is this what you meant? Also in the future please use Mathjax as it makes your question much better looking and readable – The Integrator Jun 30 '18 at 17:32
  • @TheIntegrator Thanks for the info I was wondering what I could use. That is what I meant, but on the second formula (P*r/n) is in parentheses. I don't know if that makes a difference. – Molly Jarrell Jun 30 '18 at 17:47
  • @MollyJarrell No, in this case the parantheses don´t make a difference. Is my answer comprehensible? – callculus42 Jun 30 '18 at 17:49
  • @MollyJarrell it does not make a difference here. – The Integrator Jun 30 '18 at 17:49

1 Answers1

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Firstly I replace $r/n$ by $t$. The LHS of the first equation is

$$\frac{P}{\frac{1-(1+t)^{-n}}{t}}$$

In order to divide $P$ by the fraction we multiply $P$ with the reciprocal.

$$\frac{P\cdot t }{1-(1+t)^{-n}}$$

Expanding the fraction by $(1+t)^{n}$

$$\frac{P\cdot t \cdot (1+t)^{n} }{(1+t)^{n}-1}$$

callculus42
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    instead of saying expanding i think multiplying both the numerator and denominator by $(1+t)^n$ would be better – The Integrator Jun 30 '18 at 17:51
  • @TheIntegrator Thanks for your comment. In my opinion using expanding fraction is not a bad wording. After your comment I googled it. – callculus42 Jun 30 '18 at 17:58
  • I feel so stupid. Do the negatives on the exponent cancel out after multiplying by the reciprocal? Also how does the one on the denominator move from one side to the other. – Molly Jarrell Jun 30 '18 at 18:05
  • @MollyJarrell $\frac{P\cdot t }{1-(1+t)^{-n}}\cdot\frac{(1+t)^n}{(1+t)^n}= \frac{P\cdot t \cdot (1+t)^{n} }{(1+t)^n-(1+t)^n\cdot(1+t)^{-n}}=\frac{P\cdot t \cdot (1+t)^{n} }{(1+t)^{n}-1}$ – The Integrator Jun 30 '18 at 18:10
  • @MollyJarrell First question: If you divide a term by a fraction you can equivalently multiply the term with the reciprocal. $\frac{P}{\frac{1-(1+t)^{-n}}{t}}=P\cdot \frac{t}{1-(1+t)^{-n}}=\frac{P\cdot t}{1-(1+t)^{-n}}$. Second question: Every summand at the numerator and the denominator has to be multiplied by $(1+t)^{n}$ if we expand the fraction: $\frac{P\cdot t\cdot (1+t)^{n} }{1\cdot (1+t)^{n} -(1+t)^{-n}\cdot (1+t)^{n} }=\frac{P\cdot t\cdot (1+t)^{n} }{1\cdot (1+t)^{n} -(1+t)^{-n+n} }=\frac{P\cdot t\cdot (1+t)^{n} }{ (1+t)^{n} -\underbrace{(1+t)^{0}}_{1} }$ – callculus42 Jun 30 '18 at 18:23
  • "Expanding" means "making something bigger". That's not what you are doing here: the value of the fraction is still the same. The self-contradictory description given e.g at Wyzant ("makes it bigger"/"does not change") should make it clear that "expanding" is the wrong word. – NickD Jun 30 '18 at 21:56
  • @NickD I can follow your reasoning. Feel free to edit my answer. The funny thing is that in german we use "erweitern" for such an operation. And "erweitern" has the meaning to enlarge, to expand, to increase. – callculus42 Jul 01 '18 at 02:49
  • @NickD I´m not the only one who uses this word: 1 2 3 – callculus42 Jul 01 '18 at 03:13
  • Ref. 1 is (I'm guessing from the name) by another German speaker, ref. 3 is by a Czech speaker (maybe the German terminology is used in Czech schools, but I don't know for sure). Ref.2 refers to something completely different. In any case, AFAIK, it's not common terminology in English (either British or American English). Now I'll have to go and research it in other languages :-) – NickD Jul 01 '18 at 21:59